A uniform meter scale is supported at its 20 | vedclass

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(B) Let $M$ be the mass of the meter scale and $m$ be the mass of the body. The center of mass of the uniform meter scale is at the $50 \ cm$ mark. Th

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$6$

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(B) Let $M$ be the mass of the meter scale and $m$ be the mass of the body. The center of mass of the uniform meter scale is at the $50 \ cm$ mark. The pivot $P$ is at the $20 \ cm$ mark.Case $1$: The body is at the $10 \ cm$ mark. The distance of the body from the pivot is $20 \ cm - 10 \ cm = 10 \ cm$. The distance of the center of mass of the scale from the pivot is $50 \ cm - 20 \ cm = 30 \ cm$. For rotational equilibrium,the net torque about the pivot must be zero:$Mg 	imes 30 = mg 	imes 10$$30M = 10m \Rightarrow m = 3M$Case $2$: The body is immersed in water and shifted to the $8 \ cm$ mark. The distance of the body from the pivot is $20 \ cm - 8 \ cm = 12 \ cm$. The buoyant force $F_b$ acts upwards at the $8 \ cm$ mark. The weight of the body $mg$ acts downwards at the $8 \ cm$ mark. The effective downward force at the $8 \ cm$ mark is $(mg - F_b)$.For equilibrium:$Mg 	imes 30 = (mg - F_b) 	imes 12$$30Mg = 12mg - 12F_b$Since $F_b = V ho_w g = (m / ho) ho_w g = m g / ho_r$,where $ho_r$ is the specific gravity:$30Mg = 12mg - 12(mg / ho_r)$Substitute $m = 3M$:$30Mg = 12(3M)g - 12(3M)g / ho_r$$30 = 36 - 36 / ho_r$$36 / ho_r = 6$$ho_r = 36 / 6 = 6$

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